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Topic: Inverting elliptic integrals
Replies: 6   Last Post: Aug 10, 2004 3:59 AM

 Messages: [ Previous | Next ]
 David W. Cantrell Posts: 3,395 Registered: 12/3/04
Re: Inverting elliptic integrals
Posted: Aug 9, 2004 3:08 PM

mathar@mpia.de wrote:
> In article <muqqag9mwbv2@legacy>,
> David Powers <powers@computer.org> writes:

> >I need the inverse of the (incomplete) elliptic integral of the second
> >kind. It is well defined but I can't track down a solution or code
> >for it.
> >....

>
> For a numerical representation of the inverse in terms of the angle phi,
> where E(phi,m)=int(theta=0..phi) sqrt(1-m*sin^2 theta) dtheta is the
> elliptic integral fo the second kind, one could expand E(phi,m) in a
> power series around phi=0,
>
> Emphi := phi
> -1/6*m*phi^3
> +1/5*(1/6*m-1/8*m^2)*phi^5
> +1/7*(-1/45*m+1/12*m^2-1/16*m^3)*phi^7
> +1/9*(1/630*m-1/40*m^2+1/16*m^3-5/128*m^4)*phi^9 ...
>
> The expansion coefficients in front of the order phi^(n+1) (n=2,4,6,...)
> are [sum over k=2,4,..,n of U(k,n)*F(k,m)/k!]/(n+1)!
> where
> U(k,n) = (-1)^[(k+n)/2]/2^(k-n)*[sum l=0,1,...k of (-1)^l
> (l-k/2)^n*binom(k,l)] and
> F(k,m) = -m^(n/2)*[(k-1)!!]^2/(k-1) ,
> with (k-1)!! = 1*3*5*7*...*(k-1) .
>
> Then invert this as outlined in chapt 3.6.25 of the book edited by M
> Abramowitz and I Stegun:
>
> phi := E(phi,m)
> +1/6*m*E(phi,m)^3
> +1/120*m*(13*m-4)*E(phi,m)^5
> +1/5040*m*(493*m^2-284*m+16)*E(phi,m)^7
> +1/362880*m*(37369*m^3-31224*m^2+4944*m-64)*E(phi,m)^9 ...

Yes. That's the same series I posted last week. See my Out[2], in which z
is the same as your E(phi,m).

Of course, the coefficients of the polynomials in m are obtainable by
reversion of series, as we both did. But does anyone know of a simpler way
of calculating them? (I may submit the sequence of coefficients to the OEIS
in the near future.)

Regards,
David Cantrell

Date Subject Author
3/12/97 Gordon Inverarity
7/28/04 David Powers
7/28/04 wes
7/29/04 David W. Cantrell
8/9/04 Richard Mathar
8/9/04 David W. Cantrell
8/10/04 Richard Mathar